{"id":47515,"date":"2019-11-05T06:56:49","date_gmt":"2019-11-05T05:56:49","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/exemplo-de-forca-de-arrasto-coeficiente-de-arrasto-definicao\/"},"modified":"2020-01-23T15:18:22","modified_gmt":"2020-01-23T14:18:22","slug":"exemplo-de-forca-de-arrasto-coeficiente-de-arrasto-definicao","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/pt-br\/exemplo-de-forca-de-arrasto-coeficiente-de-arrasto-definicao\/","title":{"rendered":"Exemplo de for\u00e7a de arrasto &#8211; Coeficiente de arrasto &#8211; Defini\u00e7\u00e3o"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">Exemplo: For\u00e7a de arrasto &#8211; Coeficiente de arrasto.\u00a0Calcule o atrito de uma \u00fanica barra de combust\u00edvel dentro de um n\u00facleo do reator em opera\u00e7\u00e3o normal (taxa de fluxo de projeto).\u00a0Problema com solu\u00e7\u00e3o.\u00a0Engenharia T\u00e9rmica<\/div>\n<\/div>\n<div class=\"su-divider su-divider-style-dotted\"><\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-spacer\"><\/div>\n<h2>For\u00e7a de arrasto na engenharia nuclear<\/h2>\n<p>A an\u00e1lise da\u00a0<strong>for\u00e7a de eleva\u00e7\u00e3o hidr\u00e1ulica<\/strong>\u00a0\u00e9 uma das an\u00e1lises mais importantes no projeto de um conjunto de combust\u00edvel e na an\u00e1lise da compatibilidade hidr\u00e1ulica de n\u00facleos mistos.\u00a0As for\u00e7as verticais s\u00e3o induzidas pelo fluxo ascendente de alta velocidade atrav\u00e9s do\u00a0<a title=\"N\u00facleo do reator\" href=\"https:\/\/www.reactor-physics.com\/what-is-reactor-core-definition\/\">n\u00facleo<\/a>\u00a0do\u00a0<a title=\"Reactor Core\" href=\"https:\/\/www.reactor-physics.com\/what-is-reactor-core-definition\/\">reator<\/a>\u00a0.\u00a0O caminho de fluxo para o fluido de refrigera\u00e7\u00e3o do\u00a0<a title=\"Embarca\u00e7\u00e3o de press\u00e3o do reator\" href=\"https:\/\/www.nuclear-power.com\/nuclear-power-plant\/nuclear-reactor\/reactor-pressure-vessel\/\">reator<\/a>\u00a0atrav\u00e9s do\u00a0<a title=\"Reactor Pressure Vessel\" href=\"https:\/\/www.nuclear-power.com\/nuclear-power-plant\/nuclear-reactor\/reactor-pressure-vessel\/\">vaso<\/a>\u00a0do\u00a0<a title=\"Reactor Pressure Vessel\" href=\"https:\/\/www.nuclear-power.com\/nuclear-power-plant\/nuclear-reactor\/reactor-pressure-vessel\/\">reator<\/a>\u00a0seria:<\/p>\n<ul>\n<li>\n<figure id=\"attachment_14206\" class=\"wp-caption alignright\" aria-describedby=\"caption-attachment-14206\"><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Continuity-Equation-Flow-Rate-min.png\"><img loading=\"lazy\" class=\"size-medium wp-image-14206 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Continuity-Equation-Flow-Rate-min-300x300.png\" alt=\"Equa\u00e7\u00e3o de continuidade - taxas de fluxo atrav\u00e9s do reator\" width=\"300\" height=\"300\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Continuity-Equation-Flow-Rate-min-300x300.png\" \/><\/a><figcaption id=\"caption-attachment-14206\" class=\"wp-caption-text\">Exemplo de vaz\u00f5es em um reator.\u00a0\u00c9 um exemplo ilustrativo, os dados n\u00e3o representam nenhum projeto de reator.<\/figcaption><\/figure>\n<p>O l\u00edquido de refrigera\u00e7\u00e3o entra no\u00a0<strong>vaso<\/strong>\u00a0do\u00a0<strong>reator<\/strong>\u00a0no bocal de entrada e bate no\u00a0<a title=\"Barril de n\u00facleo\" href=\"https:\/\/www.nuclear-power.com\/nuclear-power-plant\/nuclear-reactor\/core-barrel\/\">barrilete<\/a>\u00a0.<\/li>\n<li>O barrilete for\u00e7a a \u00e1gua a fluir para baixo no espa\u00e7o entre a parede do vaso do reator e o\u00a0<a href=\"https:\/\/www.nuclear-power.com\/nuclear-power-plant\/nuclear-reactor\/core-barrel\/\">barrilete<\/a>\u00a0, esse espa\u00e7o \u00e9 geralmente conhecido como\u00a0<strong>descendente<\/strong>\u00a0.<\/li>\n<li>Do fundo do vaso de press\u00e3o, o fluxo \u00e9 revertido atrav\u00e9s do n\u00facleo para passar pelos\u00a0<a href=\"https:\/\/www.nuclear-power.com\/nuclear-power-plant\/nuclear-fuel\/\">conjuntos de combust\u00edvel<\/a>\u00a0, onde a temperatura do l\u00edquido de arrefecimento aumenta \u00e0 medida que passa pelas barras de combust\u00edvel.<\/li>\n<li>Finalmente, o l\u00edquido de arrefecimento do reator mais quente entra na regi\u00e3o interna superior, onde \u00e9 direcionado pelo bico de sa\u00edda para as pernas quentes do circuito prim\u00e1rio e passa para os\u00a0<strong><a title=\"Gerador de vapor\" href=\"https:\/\/www.nuclear-power.com\/steam-generator\/\">geradores de vapor<\/a>\u00a0.<\/strong><\/li>\n<\/ul>\n<p>Os conjuntos de combust\u00edvel s\u00e3o mantidos pelo\u00a0<strong>conjunto da estrutura da guia superior<\/strong>\u00a0, que define a parte superior do n\u00facleo.\u00a0Este conjunto \u00e9 feito de a\u00e7o inoxid\u00e1vel e tem muitos prop\u00f3sitos.\u00a0O conjunto da estrutura da guia superior exerce uma\u00a0<strong>for\u00e7a axial<\/strong>\u00a0nos conjuntos de combust\u00edvel (atrav\u00e9s de molas no bico superior), define assim a posi\u00e7\u00e3o exata do conjunto de combust\u00edvel no n\u00facleo.\u00a0O\u00a0flange de\u00a0<strong>montagem da estrutura da guia superior<\/strong>\u00a0\u00e9 mantido no lugar e pr\u00e9-carregado pelo flange da cabe\u00e7a de fechamento do RPV.\u00a0O conjunto da estrutura da guia superior tamb\u00e9m orienta e protege\u00a0<a title=\"Hastes de controle\" href=\"https:\/\/www.nuclear-power.com\/nuclear-power-plant\/control-rods\/\">os<\/a>\u00a0conjuntos da\u00a0<a title=\"Control Rods\" href=\"https:\/\/www.nuclear-power.com\/nuclear-power-plant\/control-rods\/\">haste de controle<\/a>\u00a0e a instrumenta\u00e7\u00e3o interna.<\/p>\n<p>A for\u00e7a descendente necess\u00e1ria do\u00a0<strong>conjunto da estrutura da guia superior<\/strong>\u00a0nos\u00a0<strong>conjuntos<\/strong>\u00a0de combust\u00edvel deve ser calculada com muito cuidado.\u00a0For\u00e7a descendente insuficiente pode resultar na\u00a0<strong>eleva\u00e7\u00e3o do conjunto de combust\u00edvel<\/strong>\u00a0; por outro lado, uma for\u00e7a descendente excessiva pode resultar em\u00a0<strong>curvatura do conjunto de combust\u00edvel<\/strong> , o que tamb\u00e9m \u00e9 inaceit\u00e1vel.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights  lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<h2><span>Exemplo: For\u00e7a de arrasto &#8211; Coeficiente de arrasto &#8211; Pacote de combust\u00edvel<\/span><\/h2>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Hydraulic-Diameter-Fuel-Channel.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-20407 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Hydraulic-Diameter-Fuel-Channel-254x300.png\" alt=\"Di\u00e2metro hidr\u00e1ulico - canal de combust\u00edvel\" width=\"254\" height=\"300\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Hydraulic-Diameter-Fuel-Channel-254x300.png\" \/><\/a><span>Calcule o\u00a0<\/span><strong><span>atrito<\/span><\/strong><span>\u00a0de uma\u00a0<\/span><strong><span>\u00fanica barra de combust\u00edvel<\/span><\/strong><span>\u00a0\u00a0dentro de um n\u00facleo do reator em opera\u00e7\u00e3o normal (taxa de fluxo de projeto).\u00a0Suponha que essa barra de combust\u00edvel fa\u00e7a parte de um pacote de combust\u00edvel com a estrutura retangular de combust\u00edvel e esse pacote de combust\u00edvel n\u00e3o contenha grades de espa\u00e7amento.\u00a0Sua altura \u00e9\u00a0<\/span><strong><span>h = 4m<\/span><\/strong><span>\u00a0e a velocidade de fluxo do n\u00facleo \u00e9 constante e igual a\u00a0<\/span><strong><span>V\u00a0<\/span><\/strong><strong><sub><span>core<\/span><\/sub><\/strong><strong><span>\u00a0= 5 m \/ s.<\/span><\/strong><\/p>\n<p><span>Assuma isso:<\/span><\/p>\n<ul>\n<li><span>o di\u00e2metro externo do revestimento \u00e9:\u00a0<\/span><strong><span>d = 2 xr\u00a0<\/span><\/strong><strong><sub><span>Zr, 1<\/span><\/sub><\/strong><strong><span>\u00a0= 9,3 mm<\/span><\/strong><\/li>\n<li><span>o passo dos pinos de combust\u00edvel \u00e9:\u00a0<\/span><strong><span>p = 13 mm<\/span><\/strong><\/li>\n<li><span>a\u00a0<\/span><a title=\"Rugosidade relativa do tubo\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/major-head-loss-friction-loss\/relative-roughness-of-pipe\/\"><span>rugosidade relativa<\/span><\/a><span>\u00a0\u00e9\u00a0<\/span><strong><span>\u03b5 \/ D = 5 \u00d7 10\u00a0<\/span><\/strong><strong><sup><span>-4<\/span><\/sup><\/strong><\/li>\n<li><span>a\u00a0<\/span><a href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-properties\/what-is-density-physics\/\"><span>densidade<\/span><\/a><span>\u00a0do fluido\u00a0\u00e9:\u00a0<\/span><strong><span>\u03c1 = 714 kg \/ m\u00a0<\/span><\/strong><strong><sup><span>3<\/span><\/sup><\/strong><\/li>\n<li><span>a velocidade de fluxo do n\u00facleo \u00e9 constante e igual a\u00a0<\/span><strong><span>V\u00a0<\/span><\/strong><strong><sub><span>core<\/span><\/sub><\/strong><strong><span>\u00a0= 5 m \/ s<\/span><\/strong><\/li>\n<li><span>a\u00a0<\/span><a title=\"O que \u00e9 temperatura - F\u00edsica\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-temperatura-fisica-definicao\/\"><span>temperatura<\/span><\/a><span>\u00a0m\u00e9dia\u00a0do l\u00edquido de refrigera\u00e7\u00e3o do reator \u00e9:\u00a0<\/span><strong><span>T a\u00a0<\/span><\/strong><strong><sub><span>granel<\/span><\/sub><\/strong><strong><span>\u00a0= 296 \u00b0 C<\/span><\/strong><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights  lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-spacer\"><\/div>\n<h2><span>C\u00e1lculo do n\u00famero de Reynolds<\/span><\/h2>\n<p><span>Para calcular o\u00a0<\/span><a title=\"N\u00famero de Reynolds\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/reynolds-number\/\"><span>n\u00famero de Reynolds<\/span><\/a><span>\u00a0, precisamos saber:<\/span><\/p>\n<ul>\n<li><span>o di\u00e2metro externo do revestimento \u00e9:\u00a0<\/span><strong><span>d = 2 xr\u00a0<\/span><\/strong><strong><sub><span>Zr, 1<\/span><\/sub><\/strong><strong><span>\u00a0= 9,3 mm<\/span><\/strong><span>\u00a0(para calcular o di\u00e2metro hidr\u00e1ulico)<\/span><\/li>\n<li><span>o passo dos pinos de combust\u00edvel \u00e9:\u00a0<\/span><strong><span>p = 13 mm<\/span><\/strong><span>\u00a0\u00a0(para calcular o di\u00e2metro hidr\u00e1ulico)<\/span><\/li>\n<li><span>a viscosidade din\u00e2mica da \u00e1gua saturada a 300 \u00b0 C \u00e9:\u00a0<\/span><strong><span>\u03bc = 0,0000859 Ns \/ m\u00a0<\/span><\/strong><strong><sup><span>2<\/span><\/sup><\/strong><\/li>\n<li><span>a densidade do fluido \u00e9:\u00a0<\/span><strong><span>\u03c1 = 714 kg \/ m\u00a0<\/span><\/strong><strong><sup><span>3<\/span><\/sup><\/strong><\/li>\n<\/ul>\n<p><strong><span>O di\u00e2metro hidr\u00e1ulico, D\u00a0<\/span><\/strong><strong><sub><span>h<\/span><\/sub><\/strong><span>\u00a0, \u00e9 um termo comumente usado ao manipular o fluxo em\u00a0<\/span><strong><span>tubos e canais n\u00e3o circulares<\/span><\/strong><span>\u00a0.\u00a0O\u00a0<\/span><strong><span>di\u00e2metro hidr\u00e1ulico do canal de combust\u00edvel<\/span><\/strong><span>\u00a0,\u00a0<\/span><em><span>D\u00a0<\/span><\/em><em><sub><span>h<\/span><\/sub><\/em><span>\u00a0, \u00e9 igual a\u00a0<\/span><strong><span>13,85 mm<\/span><\/strong><span>\u00a0.<\/span><\/p>\n<p><span>Veja tamb\u00e9m:\u00a0<\/span><a href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/internal-flow\/hydraulic-diameter-2\/\"><span>Di\u00e2metro hidr\u00e1ulico<\/span><\/a><\/p>\n<p><span>O\u00a0<\/span><strong><span>n\u00famero de Reynolds<\/span><\/strong><span>\u00a0dentro do canal de combust\u00edvel \u00e9 ent\u00e3o igual a:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/reynolds-number-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20412 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/reynolds-number-example.png\" alt=\"n\u00famero de reynolds - exemplo\" width=\"593\" height=\"78\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/reynolds-number-example.png\" \/><\/a><\/p>\n<p><span>Isso satisfaz plenamente as\u00a0<\/span><a href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-fluxo-turbulento-definicao\/\"><strong><span>condi\u00e7\u00f5es turbulentas<\/span><\/strong><\/a><span>\u00a0.<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights  lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-spacer\"><\/div>\n<h2><span>C\u00e1lculo do coeficiente de atrito da pele<\/span><\/h2>\n<p><strong><span>O fator de atrito<\/span><\/strong><span>\u00a0para escoamento turbulento depende fortemente da\u00a0<\/span><strong><span>rugosidade relativa.\u00a0<\/span><\/strong><span>\u00c9 determinado pela equa\u00e7\u00e3o de Colebrook ou pode ser determinado usando o\u00a0<\/span><a title=\"Moody Diagram\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/major-head-loss-friction-loss\/moody-diagram\/\"><strong><span>gr\u00e1fico Moody<\/span><\/strong><\/a><span>\u00a0.\u00a0O\u00a0<\/span><strong><span>gr\u00e1fico Moody<\/span><\/strong><span>\u00a0para\u00a0<\/span><strong><span>Re = 575 600<\/span><\/strong><span>\u00a0e\u00a0<\/span><strong><span>\u03b5 \/ D = 5 x 10\u00a0<\/span><\/strong><strong><sup><span>-4<\/span><\/sup><\/strong><span>\u00a0retorna os seguintes valores:<\/span><\/p>\n<ul>\n<li><span>o\u00a0<\/span><a title=\"Fator de atrito de Darcy\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/o-que-e-o-fator-de-atrito-de-darcy-definicao\/\"><strong><span>fator de atrito de Darcy<\/span><\/strong><\/a><span>\u00a0\u00e9 igual a\u00a0<\/span><strong><span>f\u00a0<\/span><\/strong><strong><sub><span>D<\/span><\/sub><\/strong><strong><span>\u00a0= 0,017<\/span><\/strong><\/li>\n<li><span>o\u00a0<\/span><a title=\"Fator de Fric\u00e7\u00e3o\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/major-head-loss-friction-loss\/fanning-friction-factor\/\"><strong><span>fator de atrito de Fanning<\/span><\/strong><\/a><span>\u00a0\u00e9 igual a\u00a0<\/span><strong><span>f\u00a0<\/span><\/strong><strong><sub><span>F<\/span><\/sub><\/strong><strong><span>\u00a0= f\u00a0<\/span><\/strong><strong><sub><span>D<\/span><\/sub><\/strong><strong><span>\u00a0\/ 4 = 0,00425<\/span><\/strong><\/li>\n<\/ul>\n<p><span>Portanto, o coeficiente de atrito da pele \u00e9 igual a:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/skin-friction-coefficient-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20533 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/skin-friction-coefficient-example.png\" alt=\"coeficiente de atrito da pele - exemplo\" width=\"377\" height=\"186\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/skin-friction-coefficient-example.png\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights  lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-spacer\"><\/div>\n<h2><span>C\u00e1lculo da for\u00e7a de arrasto<\/span><\/h2>\n<p><span>Para calcular a\u00a0<\/span><strong><span>for\u00e7a de arrasto<\/span><\/strong><span>\u00a0, precisamos saber:<\/span><\/p>\n<ul>\n<li><span>o coeficiente de atrito da pele, que \u00e9:\u00a0<\/span><strong><span>C\u00a0<\/span><\/strong><strong><sub><span>D, atrito<\/span><\/sub><\/strong><strong><span>\u00a0= 0,00425<\/span><\/strong><\/li>\n<li><span>a \u00e1rea da superf\u00edcie do pino, que \u00e9:\u00a0<\/span><strong><span>A = \u03c0.dh = 0,11169 m\u00a0<\/span><\/strong><strong><sup><span>2<\/span><\/sup><\/strong><\/li>\n<li><span>a\u00a0<\/span><a href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-properties\/what-is-density-physics\/\"><span>densidade<\/span><\/a><span>\u00a0do fluido\u00a0, que \u00e9:\u00a0<\/span><strong><span>\u03c1 = 714 kg \/ m\u00a0<\/span><\/strong><strong><sup><span>3<\/span><\/sup><\/strong><\/li>\n<li><span>a velocidade de fluxo do n\u00facleo, que \u00e9 constante e igual a\u00a0<\/span><strong><span>V\u00a0<\/span><\/strong><strong><sub><span>core<\/span><\/sub><\/strong><strong><span>\u00a0= 5 m \/ s<\/span><\/strong><\/li>\n<\/ul>\n<p><span>A partir do\u00a0<\/span><strong><span>coeficiente de atrito<\/span><\/strong><span>\u00a0da\u00a0<strong>pele,<\/strong>\u00a0que \u00e9 igual ao\u00a0<\/span><a href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/major-head-loss-friction-loss\/fanning-friction-factor\/\"><strong><span>fator de atrito de Fanning<\/span><\/strong><\/a><span>\u00a0, podemos calcular o componente de atrito da\u00a0<\/span><strong><span>for\u00e7a de arrasto.\u00a0<\/span><\/strong><span>A for\u00e7a de arrasto \u00e9 dada por:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/drag-force-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20534 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/drag-force-example.png\" alt=\"for\u00e7a de arrasto - exemplo\" width=\"452\" height=\"72\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/drag-force-example.png\" \/><\/a><\/p>\n<p><span>Supondo que um conjunto de combust\u00edvel possa ter, por exemplo, 289 pinos de combust\u00edvel (conjunto de combust\u00edvel 17 \u00d7 17), o\u00a0<\/span><strong><span>componente<\/span><\/strong><span>\u00a0de\u00a0<strong>atrito<\/strong>\u00a0da\u00a0<\/span><strong><span>for\u00e7a de arrasto<\/span><\/strong><span>\u00a0\u00e9 da ordem de\u00a0<\/span><strong><span>quilonewtons<\/span><\/strong><span>\u00a0.\u00a0Al\u00e9m disso, essa for\u00e7a de arrasto se origina puramente do atrito da pele no pacote de combust\u00edvel.\u00a0Mas o conjunto t\u00edpico de combust\u00edvel PWR cont\u00e9m outros componentes, que influenciam o sistema hidr\u00e1ulico do conjunto:<\/span><\/p>\n<ul>\n<li><strong><span>Barras de combust\u00edvel<\/span><\/strong><span>\u00a0.\u00a0As barras de combust\u00edvel cont\u00eam o combust\u00edvel e os venenos queimados.<\/span><\/li>\n<li><strong><span>Bico superior<\/span><\/strong><span>\u00a0.\u00a0Fornece o suporte mec\u00e2nico para a estrutura de montagem de combust\u00edvel.<\/span><\/li>\n<li><strong><span>Bico inferior<\/span><\/strong><span>\u00a0.\u00a0Fornece o suporte mec\u00e2nico para a estrutura de montagem de combust\u00edvel.<\/span><\/li>\n<li><strong><span>Grade de espa\u00e7amento<\/span><\/strong><span>\u00a0.\u00a0Garante uma orienta\u00e7\u00e3o exata das barras de combust\u00edvel.<\/span><\/li>\n<li><strong><span>Guie o tubo do dedal<\/span><\/strong><span>\u00a0.\u00a0Tubo vago para hastes de controle ou instrumenta\u00e7\u00e3o interna.<\/span><\/li>\n<\/ul>\n<p><span>Como foi escrito, o segundo componente da for\u00e7a de arrasto \u00e9 o arrasto do formul\u00e1rio.\u00a0<\/span><strong><span>O arrasto de forma<\/span><\/strong><span>\u00a0conhecido tamb\u00e9m como\u00a0<\/span><strong><span>arrasto de press\u00e3o<\/span><\/strong><span>\u00a0surge devido \u00e0\u00a0<\/span><strong><span>forma<\/span><\/strong><span>\u00a0e\u00a0<\/span><strong><span>tamanho<\/span><\/strong><span>\u00a0do objeto.\u00a0O\u00a0<\/span><strong><span>arrasto de press\u00e3o<\/span><\/strong><span>\u00a0\u00e9 proporcional \u00e0 diferen\u00e7a entre as press\u00f5es que atuam na frente e nas costas do corpo imerso e a \u00e1rea frontal.<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights  lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-spacer\"><\/div>\n<h2><span>Queda de press\u00e3o &#8211; Montagem de combust\u00edvel<\/span><\/h2>\n<p><span>Em geral,\u00a0<\/span><strong><span>a\u00a0<\/span><\/strong><strong><span>queda de press\u00e3o<\/span><\/strong><span>\u00a0total\u00a0<strong>do conjunto de combust\u00edvel<\/strong>\u00a0\u00e9 formada pela queda de atrito do feixe de combust\u00edvel (dependente da\u00a0<\/span><a href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/major-head-loss-friction-loss\/relative-roughness-of-pipe\/\"><span>rugosidade relativa<\/span><\/a><span>\u00a0das barras de combust\u00edvel,\u00a0<\/span><a href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/reynolds-number\/\"><span>n\u00famero de reynolds<\/span><\/a><span>\u00a0,\u00a0<\/span><a href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/internal-flow\/hydraulic-diameter-2\/\"><span>di\u00e2metro hidr\u00e1ulico<\/span><\/a><span>\u00a0etc.) e outras quedas de press\u00e3o dos elementos estruturais (bico superior e inferior, grades espa\u00e7adoras ou grades de mistura) )<\/span><\/p>\n<p><span>Em geral, n\u00e3o \u00e9 t\u00e3o simples calcular as quedas de press\u00e3o nos conjuntos de combust\u00edvel (especialmente as grades espa\u00e7adoras) e isso pertence ao\u00a0<\/span><strong><span>know-how<\/span><\/strong><span>\u00a0essencial\u00a0de determinado fabricante de combust\u00edvel.\u00a0Principalmente, as quedas de press\u00e3o s\u00e3o medidas em\u00a0<\/span><strong><span>la\u00e7os hidr\u00e1ulicos experimentais<\/span><\/strong><span>\u00a0, em vez de calculadas.<\/span><\/p>\n<p><span>Os engenheiros usam o\u00a0<\/span><a title=\"Coeficiente de perda de press\u00e3o - PLC\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/bernoullis-equation-bernoullis-principle\/head-loss\/pressure-loss-coefficient-plc\/\"><strong><span>coeficiente de perda de press\u00e3o<\/span><\/strong><\/a><span>\u00a0,\u00a0<\/span><strong><span>CLP<\/span><\/strong><span>\u00a0.\u00a0Nota-se K ou\u00a0<\/span><strong><span>\u03be<\/span><\/strong><span>\u00a0\u00a0(pronunciado \u201cxi\u201d).\u00a0Este coeficiente caracteriza a\u00a0<\/span><a title=\"Perda de Cabe\u00e7a - Perda de Press\u00e3o\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/bernoullis-equation-bernoullis-principle\/head-loss\/\"><span>perda<\/span><\/a><span>\u00a0de\u00a0<a title=\"Perda de Cabe\u00e7a - Perda de Press\u00e3o\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/bernoullis-equation-bernoullis-principle\/head-loss\/\">press\u00e3o<\/a>\u00a0de um determinado sistema hidr\u00e1ulico ou de uma parte de um sistema hidr\u00e1ulico.\u00a0Pode ser facilmente medido em loops hidr\u00e1ulicos.\u00a0O coeficiente de perda de press\u00e3o pode ser definido ou medido para tubos retos e especialmente para\u00a0<\/span><a title=\"Menor Perda de Cabe\u00e7a - Perdas Locais\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/minor-head-loss-local-losses\/\"><strong><span>perdas locais (menores)<\/span><\/strong><\/a><span>\u00a0.<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/PLC-Pressure-loss-coefficient-equations.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-20526 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/PLC-Pressure-loss-coefficient-equations.png\" alt=\"CLP - Coeficiente de perda de press\u00e3o - equa\u00e7\u00f5es\" width=\"339\" height=\"375\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/PLC-Pressure-loss-coefficient-equations.png\" \/><\/a><\/p>\n<p><span>Usando os dados do exemplo acima mencionado, o\u00a0<\/span><strong><span>coeficiente de perda de press\u00e3o<\/span><\/strong><span>\u00a0(apenas por atrito do tubo reto) \u00e9 igual a\u00a0<\/span><strong><span>\u03be = f\u00a0<\/span><sub><span>D<\/span><\/sub><span>\u00a0L \/ D\u00a0<\/span><sub><span>H<\/span><\/sub><span>\u00a0= 4.9<\/span><\/strong><span>\u00a0.\u00a0Mas o coeficiente geral de perda de press\u00e3o (incluindo grades espa\u00e7adoras, bocais superior e inferior etc.) \u00e9 geralmente cerca de tr\u00eas vezes maior.\u00a0Este PLC (\u00a0<\/span><strong><span>\u03be = 4,9<\/span><\/strong><span>\u00a0) faz com que a queda de press\u00e3o \u00e9 da ordem de (usando as entradas anteriores)\u00a0<\/span><strong><span>Ap\u00a0<\/span><sub><span>atrito<\/span><\/sub><\/strong><span>\u00a0= 4,9 x 714 x 5\u00a0<\/span><sup><span>2<\/span><\/sup><span>\u00a0\/2 =\u00a0<\/span><strong><span>43,7 kPa<\/span><\/strong><span>\u00a0(sem grelhas de espa\u00e7amento, de topo e bocais inferiores).\u00a0O CLP real cerca de tr\u00eas vezes maior significa que o\u00a0<strong><sub>combust\u00edvel\u00a0<\/sub><\/strong><\/span><strong><span>\u00a0\u0394p<\/span><\/strong><span>\u00a0cerca de tr\u00eas vezes maior\u00a0ser\u00e1.<\/span><\/p>\n<p><span>A perda de press\u00e3o geral do reator,\u00a0<strong><sub>reator<\/sub><\/strong><\/span><strong><span>\u00a0\u0394p<\/span><\/strong><span>\u00a0, deve incluir:<\/span><\/p>\n<ul>\n<li><span>descendente e fundo do reator<\/span><\/li>\n<li><span>placa de suporte inferior<\/span><\/li>\n<li><span>conjunto de combust\u00edvel, incluindo grades espa\u00e7adoras, bocais superior e inferior e outros componentes estruturais &#8211;\u00a0\u00a0<\/span><strong><span>\u0394p\u00a0<\/span><sub><span>fuel<\/span><\/sub><\/strong><\/li>\n<li><span>montagem da estrutura da guia superior<\/span><\/li>\n<\/ul>\n<p><span>Como resultado, a perda de press\u00e3o total do reator &#8211;\u00a0\u00a0<\/span><strong><span>\u0394p do\u00a0<\/span><sub><span>reator<\/span><\/sub><\/strong><span>\u00a0\u00e9 geralmente da ordem de centenas de kPa (digamos 300 &#8211; 400 kPa) para par\u00e2metros de projeto.<\/span><\/p>\n<\/div>\n<\/div>\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.<\/p>\n<p>Este artigo \u00e9 baseado na tradu\u00e7\u00e3o autom\u00e1tica do artigo original em ingl\u00eas. Para mais informa\u00e7\u00f5es, consulte o artigo em ingl\u00eas. Voc\u00ea pode nos ajudar. Se voc\u00ea deseja corrigir a tradu\u00e7\u00e3o, envie-a para: translations@nuclear-power.com ou preencha o formul\u00e1rio de tradu\u00e7\u00e3o on-line. Agradecemos sua ajuda, atualizaremos a tradu\u00e7\u00e3o o mais r\u00e1pido poss\u00edvel. Obrigado.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Exemplo: For\u00e7a de arrasto &#8211; Coeficiente de arrasto.\u00a0Calcule o atrito de uma \u00fanica barra de combust\u00edvel dentro de um n\u00facleo do reator em opera\u00e7\u00e3o normal (taxa de fluxo de projeto).\u00a0Problema com solu\u00e7\u00e3o.\u00a0Engenharia T\u00e9rmica For\u00e7a de arrasto na engenharia nuclear A an\u00e1lise da\u00a0for\u00e7a de eleva\u00e7\u00e3o hidr\u00e1ulica\u00a0\u00e9 uma das an\u00e1lises mais importantes no projeto de um conjunto &#8230; <a title=\"Exemplo de for\u00e7a de arrasto &#8211; Coeficiente de arrasto &#8211; Defini\u00e7\u00e3o\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/pt-br\/exemplo-de-forca-de-arrasto-coeficiente-de-arrasto-definicao\/\" aria-label=\"More on Exemplo de for\u00e7a de arrasto &#8211; Coeficiente de arrasto &#8211; Defini\u00e7\u00e3o\">Ler mais<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[14],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v15.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Exemplo de for\u00e7a de arrasto - Coeficiente de arrasto - Defini\u00e7\u00e3o<\/title>\n<meta name=\"description\" content=\"Exemplo: For\u00e7a de arrasto - Coeficiente de arrasto. 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Calcule o atrito de uma \\u00fanica barra de combust\\u00edvel dentro de um n\\u00facleo do reator em opera\\u00e7\\u00e3o normal (taxa de fluxo de projeto). Problema com solu\\u00e7\\u00e3o. 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